Organic Chemistry Enols and Enolates, Part 1

Instead of having an OH group, Carboxylic Acid Derivatives can have a halide, anhydride, esters, amides, and nitriles. These groups all have lone pairs and can act as leaving groups which can be ranked by electrophilicity due to inductive and resonance effects. These groups can be prepared by interconversion as well as derivative-specific syntheses starting from a carboxylic acid or other carbonyl. Esters in particular have several syntheses, including Fischer esterification, transesterification, and SN2 with a carboxylate using diazomethane. Amides, however, cant be directly synthesized from carboxylic acids very easily because amines are basic and will deprotonate the acid. Carboxylic Acid Derivatives react with nucleophiles and organometallic reagents, and they can be reduced using hydrides like Lithium Aluminum Hydride.

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I download all power point and print out to study but why the slide order dose not match with lecture slide order? I am just so confused. How can do arrange these slide order?

2 answers

Last reply by: Some oneWed Apr 24, 2013 4:17 AM

Post by Some oneon April 23, 2013

Hi professor, we mentioned that it usually takes strong nucleophiles to attack the carbonyl carbon, such as LAH and Grignard reagents. What makes the enolate such a strong nucleophile in the aldol reaction that it would disrupt the resonance stabilized carbonyl. Thank you so much.

1 answer

Last reply by: Professor StarkeyMon Apr 22, 2013 6:55 PM

Post by Some oneon April 22, 2013

Hello professor, why is C-alkylation preferred over O-alkylation on an enolate?

1 answer

Last reply by: Professor StarkeySun Mar 31, 2013 12:35 PM

Post by Marrbell Marteyon March 30, 2013

What would the product be if aldehyde is used in Dibenzalacetone formation

1 answer

Last reply by: Professor StarkeySun Apr 29, 2012 9:38 AM

Post by Rachel Paquetteon April 27, 2012

at 41:00 you are explaining the mechanism for the aldol condensation, and you say that the alpha hydrogen will be deprotonated because that is the theme but if you have OH wouldn't you be able to attack the carbonyl carbon and then move the electrons in the double bond move up to the oxygen? how do you know whether you are deprotonating the alpha hydrogen or attacking the carbonyl carbon

0 answers

Post by Misael Nietoon March 28, 2012

I love the detail in this lectures. Thankyou

1 answer

Last reply by: Professor StarkeyMon Jan 23, 2012 11:29 PM

Post by Jason Jarduckon January 22, 2012

Hi Dr. Starkey,

I like your lecture alot!! Today I watched the complete lecture to find the answer to a question.

Thank you

Jason Jarduck

0 answers

Post by Jamie Spritzeron August 12, 2011

in 15:49, where is the counter-ion Na+ going to be? If there are 2 resonance forms, is it with one O- or the other one, depending on the resonance form?

1 answer

Last reply by: Professor StarkeyThu Oct 27, 2011 11:02 AM

Post by Jamie Spritzeron August 10, 2011

In 65:22 the base LDA is used to deprotonate, but in the second step at the end, there is an acidic workup with H3o+. How can you have both acid and base in the same mechanism?

1 answer

Last reply by: Professor StarkeyThu Oct 27, 2011 11:05 AM

Post by Jamie Spritzeron August 7, 2011

Is the acidity of the alpha proton in a carboxylic acid about the same as for an ester?

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Enols and Enolates, Part 1

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Transcription: Enols and Enolates, Part 1

Today, we are going to talk about enols and enolates, which are components of aldehydes and ketones.0003

So far, what we have seen for aldehydes and ketones involved the reactivity of the carbonyl, the C-O double bond; and the most significant part that we have seen so far is the polarity of the C-O double bond due to the resonance, which makes this carbon partially positive and makes it an electrophile.0010

In other words, nucleophiles add here; and that is certainly a significant reactivity of aldehydes and ketones--of any carbonyl.0030

What we are going to shift toward in this lesson is looking at the alpha carbon (that is the carbon right next to the carbonyl): and consider the protons that are attached to that carbon.0041

It turns out that these protons, the alpha protons, are acidic: and what does it mean to be an acid?--it means you donate a proton.0052

In this lesson, we are going to explore the deprotonation of the α carbon and the consequences of that deprotonation.0069

One other component that is of interest to us in this unit is the idea that any carbonyl-containing compound (like a ketone or an aldehyde) is in equilibrium with another form called an enol.0077

Now, this structure is called an enol because it has an alkene (a carbon-carbon double bond) and an alcohol (OH), and they are on the same carbon.0093

The interconversion between a ketone and its enol form is known as a tautomerization, and as you can see, the equilibrium is favored in the reverse direction here: the ketone is more stable.0104

Having that carbonyl there is preferable, so that is the form that it prefers to be, but what is important to recognize is that there is always a small amount present here of the enol--a small amount present at equilibrium.0117

So, in other words, any time you have a ketone or an aldehyde or any carbonyl-containing compound, you also have some of the enol form around; that is going to be important to us when developing some of the mechanisms, down the road.0136

Let's look, first, at this tautomerization mechanism: how do we go from a ketone to an enol?0149

First, let's draw the enol form, because that is going to be helpful for us to get from one place to the other.0155

Let's think about where we are headed; and when we compare these two structures, we see that there are two things that need to be accomplished, and that is why we have a two-step mechanism here.0165

It looks like we have an oxygen here, and now it's an OH; so one of my steps is: I need to protonate here at this oxygen--I need to add a proton at some point.0175

And here we have a CH3, and now it's a CH2, so I need to deprotonate down here.0187

And, of course, the π bond has moved as well; that is going to be part of our mechanism, and could be seen as resonance, too.0195

The only question we have is: those are our two steps--what step are we going to do first?0203

Well, typically, this tautomerization--the presence of enol is going to be something that we find in acidic conditions, so in acidic conditions, let's protonate first; so that is going to be our step 1 (protonate), and step 2 is going to be deprotonate.0207

So, I can protonate; let's add in our lone pairs on this oxygen: step 1 will be "protonate the carbonyl oxygen."0220

That is going to give an oxygen with three bonds and a lone pair: 1, 2, 3, 4, 5; that is a positively charged oxygen.0239

OK, and my second step, then, is going to be to deprotonate this α carbon.0248

It turns out we are deprotonating right away; so instead of making a CH3, let's make this a CH2, so that we can see one of the hydrogens involved.0255

What we want to do is deprotonate that hydrogen; we can use the A- that we formed in this first step to be a base; and can you see how the mechanism might work to get us to our enol structure?0263

Typically, when you do a deprotonation, the electrons just go and sit on this atom; but instead of moving them here, we can move them over to be a π bond, which would push this π bond up onto the oxygen; and now, we end up with the best resonance form here, where we have no formal charges.0280

Any time we need to do a tautomerization, it is going to be a two-step mechanism; and we should practice this both forward and backwards until we get pretty good at it, so that it will be automatic when it comes to part of a larger mechanism.0298

This is now the more favorable mechanism that we will see, and what do we have to do?0321

Well, again, it is still going to be two steps: the reverse mechanism (we already have it written down here) is going to be the same number of steps and involve the same intermediates as the forward mechanism.0328

And so, what do we need to accomplish?--well, we have an OH that now needs to go back to an oxygen; so one thing I need to do is: I need to deprotonate.0339

It really helps to think about where you are headed and what you have to accomplish before you dive into a mechanism.0350

And my CH2 becomes a CH3, so I am going to protonate in this position.0355

Those are going to be my two steps: and what do you think we should do first?0364

Well, I am thinking again, because we are in acidic conditions, we should protonate first; so that should always be our first step in our tautomerization.0367

Let's redraw this: I need to protonate at that CH2; so I can bring an acid over here, HA, and how could I protonate this carbon?0375

Well, I can use this π bond as my base; I can grab that proton, and that way, I can add a proton to that position.0388

What happens to this carbon now: it only has three bonds, and so I have a positive charge.0401

Just like any time I protonate a π bond, like we saw for alkenes, you get a carbocation as your product.0405

Our second step (that was step 1: to protonate): step 2 is going to be to deprotonate, so I can bring my A- in here to deprotonate this oxygen.0412

Once again, rather than deprotonate and give an O- next to a C+, my mechanism would be much more efficient if I grabbed that proton and took these two electrons and filled them in as a π bond between the carbon and the oxygen.0423

There we go--our two steps, and we have gone from our enol to our ketone.0442

Now, I mentioned at the beginning that we are going to be able to deprotonate α carbons, and let's think about why a ketone's α protons would be acidic.0449

As usual, any time we want to determine something about the acidity of a compound, we should take a look at the conjugate base, and see how stable that is.0459

Here is an α proton; the pKa for such a proton is on the order of about 20, and if we imagine deprotonating it (treating it with some strong base that can come in and remove it), we would get this carbanion.0466

Now, normally a carbanion is not a very happy charge, because carbon doesn't like having a negative charge; but there is something about this conjugate base that makes it reasonably stable.0483

Well, that is because the lone pair is allylic; it is next to a π bond; in fact, it is next to a carbonyl, which means it has resonance; I can delocalize this charge.0494

Any time I have an allylic lone pair, I can bring the π bond in, and bring this π bond up, and I can draw a second resonance form.0506

Any time I could draw a second resonance form, that is always a good thing; and that explains why this is a reasonable place to deprotonate, because the conjugate base is resonance-stabilized.0516

It is a resonance-stabilized carbanion; so it is a carbanion, but it is stabilized by resonance.0533

And we can not only delocalize that charge, but we could delocalize that charge to the more electronegative oxygen; so this is an excellent resonance form.0543

Now, there is a name for this resonance-stabilized carbanion: it is called an enolate.0550

We just saw what an enol looks like: an enol has a double bond with an OH attached to that, right on that double bond; an enolate is the charged version of that structure.0555

And so, as you could see, the title of this lesson is "Enols and Enolates"; so those are going to be the two species that we are going to be interested in both forming and reacting.0569

It turns out that we are going to be using both the enol and the enolate; if we have acid-catalyzed mechanisms, that is when we will be using the enol version of this compound, and in a base-catalyzed mechanism is when we are going to be using the enolate version.0586

Now, before we progress, though, I want to talk a little bit about deprotonating a carbon; that is not a very easy thing to do.0607

We are going to see repeatedly, throughout this unit, that we can do it for α carbons; let's think about other types of CH's.0614

OK, if you have an alkyne, that has a reasonable pKa, as well--somewhere around 25; that means...that is a low enough number that, if you do treat it with a strong base, it is possible to deprotonate here and form this carbanion.0623

Remember, this is sp hybridized, and that is what made this possible; and this is an OK carbanion to make.0637

We have made these; we have used these in synthesis; and this is something that you should recognize as a staple carbanion.0647

OK, however, if you tried to do that same thing with an alkane or an alkene--if you try to take an sp3 hybridized carbon or an sp2 hybridized carbon and try and do the same thing and react with a base, you would get a completely unstable carbanion here and a completely unstable carbanion here.0654

We cannot deprotonate alkanes and alkenes; OK, it is possible to deprotonate alkynes, and the focus of this unit, of course, is deprotonating α carbons, such as aldehydes and ketones.0674

Now, if we are going to be protonating, that means we need a base; let's think about what bases are possible.0689

If you had to name a base, you might say something like sodium hydroxide; we know that is a good base, so let's try that.0693

Let's react this with sodium hydroxide and think about what product we would get.0699

OK, I could use hydroxide here as my base; it can grab this proton; and rather than have the electrons sit right on carbon, most often, when we deprotonate an enolate, we are going to go right to that better resonance form.0707

We are going to take these electrons and form a π bond and move those electrons all the way up to be on that oxygen.0723

That is the preferred resonance form--that is the better resonance form.0729

Water is the stronger acid; and how does the equilibrium lie in a proton transfer reaction?--it always lies in the direction of the weaker acid-base pair.0761

So, this reaction can happen, but just to a small extent; it is actually the reverse reaction that is favored in this case.0772

So, if I were to use a base like sodium hydroxide, what would happen is: I would deprotonate just a small amount of this ketone; I would make a small amount of the enolate; a small amount of enolate will be present at equilibrium.0780

But, for the most part, the ketone would still be a ketone; it would be the neutral form.0802

This is exactly the situation we want in certain reactions; so in those cases, we use sodium hydroxide.0808

We also want to recognize--let's just make that note here--we have a large amount of the neutral ketone or aldehyde present at equilibrium.0817

So, with a weak base like hydroxide, we have mostly ketone with a tiny little bit of enolate around.0830

If we wanted to completely deprotonate the ketone or aldehyde to make it an enolate, 100%, we are going to need a much stronger base.0837

One of the bases that has been developed for this, that is used very widely, is this one; this is called lithium diisopropylamide (that means we have an N-).0846

Lithium diisopropylamide is very conveniently abbreviated as LDA; and this is a great base, because it is very strong base, and it is very bulky (having those two isopropyl groups introduces a lot of steric hindrance).0858

So, in other words, LDA is just a base; all it does is find a proton to take, and that is its only purpose.0882

Where do we have acidic protons?--of course, the α protons are acidic, so we can deprotonate here.0890

Again, we could show that mechanism; the N- can come and be the base, and we can make this enolate.0897

This would be the lithium enolate; it would be the lithium cation here, because it's lithium diisopropylamide.0907

We most often don't have the counterion there, but of course, it's there every time we have a negative charge.0914

OK, but the other product here, now, is going to be diisopropylamine: diisopropylamine has a pKa of around 40--this is a very, very weak acid.0920

That means this reverse reaction of having an enolate try and deprotonate diisopropylamine is very, very small: essentially nonexistent.0930

We can think of this as a one-way street--as a way to completely deprotonate a ketone or an aldehyde and convert it to its enolate form.0939

Now, we don't always have to use LDA to do a complete deprotonation; there are cases, like this one, known as an active methylene; an active methylene is when you have this situation, where this carbon is between two carbonyls.0951

That makes it even more acidic: the pKa...let's compare it to this α carbon (this α carbon has just one carbonyl, so it is similar to an ordinary ketone): this has a pKa somewhere around 20.0971

OK, but when you put one next to 2 carbonyls, you have a pKa somewhere around 11.0986

When you think about deprotonating anywhere else, neither of these is at all acidic, being next to an oxygen or just being a plain old alkyl; remember, we are never going to deprotonate an ordinary alkyl hydrogen, so these are not acidic at all.0997

It is only the ones that are going to be α to carbonyls that we are going to be considering deprotonating in this unit.1016

Sodium methoxide is a good base; so we could use that as our base, and if we did that, we would (let's just draw a line drawing here) deprotonate this completely.1021

Again, the reverse reaction is minimal; we can completely deprotonate this α proton, and that is because this enolate not only has the carbanion (and we could delocalize it into this carbon on this oxygen on the left--the carbonyl on the left), but this could also have resonance to the other carbonyl, and that is what makes this conjugate base so very stable, and therefore makes the parent acid so acidic.1038

OK, my other product here--if I used ethoxide, it gives me ethanol; that has a pKa of somewhere around 18.1070

So again, comparing a pKa of 11 to a pKa of 18, that is a huge difference, and that strongly favors the forward direction and negates the reverse direction.1079

This is called a stabilized enolate: when we have something that is next to two carbonyls that resulting enolate has extra resonance, and so we describe it as a stabilized enolate.1091

In this case, you could use LDA; but as usual, we always want to use a reagent that is the lowest reactivity, most stable; that makes it easier to handle and less expensive.1108

And so, sodium ethoxide would work just as well here, so we would use that.1118

Just a quick question: we talked about how sodium hydroxide is something that can be used to form a small amount of enolate: why not use sodium hydroxide in this case?1122

That certainly is on the order of the same basicity as ethoxide, so it would also do a good job of deprotonating this.1133

But is there another reaction that can happen if I use sodium hydroxide as a base instead of sodium ethoxide?1141

Well, take a look here: we have an ester in this starting material; so what would happen if you took this ester?--there is another reaction that can happen, besides just the acid-base reaction.1150

We could also have hydroxide add into the carbonyl, and then come back down and kick off the ethoxide; so we could also get hydrolysis substitution--an acyl substitution--here, if we used hydroxide.1163

Notice: because this is an ethyl ester, what base did we choose to use? we used the ethoxide.1177

OK, so the key here is: for esters, when we go to choose a base and we want to deprotonate α to an ester carbonyl, we are going to use the matching base.1186

So, if we have an O-R ester, we are going to use the alkoxide that matches that; so if it's a methyl ester, we use methoxide here; we had an ethyl ester--we used ethoxide; and that is to avoid the substitution reaction that can happen with the carbonyl.1202

If you have ethoxide, ethoxide can add into this carbonyl, but what is it going to be kicking out?1219

It is going to be replacing ethoxide; so that reaction becomes invisible, and no longer a side product.1223

Now, I just started to introduce how you could have esters be deprotonated at the α carbon; and that is true--we are focusing most of our reaction on aldehydes and ketones initially.1231

But you can have any carbonyl; so here we have a proton that is α to a carbonyl that has a pKa somewhere around 20; we can deprotonate here.1243

Here we have, again, an α proton that is next to a carbonyl, but that carbonyl happens to be an ester; that is still acidic--not quite as acidic as the ketone, but it is still acidic, and we still can deprotonate here.1253

OK, but carbonyls are not the only groups that impart acidity to the carbon adjacent to it.1266

OK, the nitro group, the NO2 group, also makes this carbon acidic--in fact, significantly more acidic than having just a carbonyl.1273

And the cyano group is another group that makes α carbons acidic, kind of like the same as an ester might.1285

OK, and let's think about...let's draw those conjugate bases to see if it makes sense to us; this should make sense on why each of these hydrogens is acidic.1294

If we have a ketone, and we put an anion here, why would that be an appropriate place to have an anion?--well, because it is next to the carbonyl, and it can have resonance.1307

OK, how about if I had an ester?--I'm sorry...I have an ester...I have a methoxy group over here, instead of just a methyl group.1322

Well, we could still have resonance here, and this is still a resonance-stabilized enolate, so you can make the enolate of an ester, just like you can for a ketone.1332

OK, but how about the nitro or the cyano?--well, let's take a look at the structure of the nitro group.1343

Let's imagine deprotonating; so we just lost a proton here, and we now have this anion.1348

When you draw out the nitro group, it looks something like this; and how would that be a stable carbanion?1356

Well, this can have resonance just like a carbonyl can; we have an N-O double bond instead of a C-O double bond; but just like the carbonyl could, we can use that to delocalize the negative charge and move it to an oxygen atom.1370

Of course, having a negative charge on an oxygen atom means this is a very good resonance form.1385

OK, plus you have this positively charged nitrogen pulling electron density and stabilizing inductively; that is why this is even more stable than an enolate; it makes the α proton so much more acidic.1390

How about a nitrile?--let's imagine this anion and draw out the nitrile: a nitrile, of course, has a C-N triple bond.1401

Well, again, we have an allylic negative charge and an allylic lone pair, so the lone pair can come in, and the π bond can move over.1411

We can have resonance; it puts the negative charge on nitrogen; that is not as good as oxygen, but it is certainly better than carbon, and it is because of this resonance stabilization that it makes it a reasonably acidic proton and something we can deprotonate with a strong base like LDA.1420

These are enolate-like species; when we say "enolate," we mean a carbonyl of some kind--either a ketone or an aldehyde or an ester; but these are enolate-like species, as well.1437

Just a quick question: there is a big difference here--5 orders of magnitude less acidic for an ester, compared to a ketone: why do you think that is less acidic--what is there about this conjugate base that is better than this conjugate base?1451

So, an ester carbonyl is more electron rich; so that means it does not want to have this negative charge--this negative charge, now, has to compete with the oxygen's electron donation for this carbonyl, so there is less resonance stabilization that can go on to delocalize that negative charge.1485

OK, so it is because the ester carbonyl is more electron rich that that makes it a weaker acid.1509

Just to summarize: the carbonyl, the nitro, the cyano...each of those groups are described as electron-withdrawing groups (or EWG for short); and what they all have in common is: if you put a negative charge right next door to them, they would be able to stabilize that negative charge by resonance.1519

That puts these three groups together in a certain category; we are going to see lots of examples, actually, where the EWG comes into play, and imparts a particular reactivity to the groups attached to it.1539

So, now that we have taken a look at how to make enolates with a mechanism, what kinds of bases we can use, what types of α protons are going to be acidic...what would we do with an enolate--how are they used?1555

Well, they are carbanions; they are examples of carbons with a negative charge, which makes them good nucleophiles: anything that is electron rich is going to be a good nucleophile.1567

Electrophiles are things that are electron poor: things like an alkyl halide--that is an electrophile we have seen: we know that this halogen (chloride, bromide, iodide...) pulls electron density from this carbon, making it partially positive.1582

So, we can have nucleophiles attack here, doing something like an SN2 mechanism, for example; and so, an enolate would be an example of that.1600

OK, or even just Br2 or a halide like that--it doesn't have a carbon bearing a leaving group, but it has a halogen bearing a leaving group.1613

For that same reason, you could have an SN2 occurring with that; we will see reactions of enolates with something like Br2.1624

Now, a carbonyl has a partial positive/partial minus; so we have seen carbonyl reactivities as electrophiles, and we will see something like a ketone, aldehyde, ester...OK, in general, any carbonyl would be a good electrophile; and we will see enolates react with those.1635

And finally, an epoxide would be a good electrophile, as well; remember, the carbons of the epoxide are partially positive, because the oxygen pulls electron density.1657

There is a lot of ring strain involved in that: it's a three-membered ring, and so those also readily react with nucleophiles.1667

And of course, it is that α proton that is going to be acidic; so step 1 is going to deprotonate the α carbon to give us a carbanion.1706

OK, and what I'm suggesting is that this carbanion, now, is going to be a very, very good nucleophile.1719

In step 2, when I react with methyl iodide (which is a good electrophile), what reaction can happen here?1726

I think the nucleophile is going to attack the carbon and kick off the leaving group; that looks like a back side attack--it looks like an SN2 mechanism.1736

That is exactly what happens, and I have now installed a methyl group at the α carbon; so this is called an alkylation or an α alkylation reaction.1744

It is this two-step process: we deprotonate, and then we treat it with an alkyl halide.1759

Now, because it is an SN2, remember, that back side attack requires minimum steric hindrance, so we have to have an unhindered alkyl halide; a methyl group would be great; primary would be fine; but as soon as we get to secondary or tertiary, then E2 is going to be preferred, because the enolate is still a pretty strong base.1764

Now, as shown here, the enolate is a nucleophile at carbon: in other words, C alkylation--putting the alkyl group on the carbon--is preferred, rather than O alkylation.1783

OK, and this is important to point out, because remember: the enolate has two resonance forms.1794

It doesn't matter which resonance form you draw; you should still be getting the same product.1800

OK, so let's take a look at the mechanism using the better resonance form, the preferred resonance form.1804

What we'll do is: we will have our base, our LDA, come in and deprotonate; but instead of putting the negative charge on this carbon, the more appropriate place to put that negative charge is on the oxygen.1810

Now, that is the better resonance form, because it has the negative charge on the more electronegative oxygen; in other words, this better represents the actual hybrid of the enolate structure.1831

However, it is very tempting now, when you look at this, to think, "When I bring this together with my electrophile, it's very tempting to use the negative charge from oxygen to go and attack."1841

That would be described as O alkylation, where the alkyl group ends up on the oxygen; that does not happen.1854

That does not happen; it is the α carbon; it is the α carbon that is the nucleophile in an enolate, no matter how you draw it; it is the α carbon.1860

So, how do I get this carbon to react?--well, this is what enolates do: the enolate is here: the enolate is a nucleophile; and I'm going to start with the lone pair up on the oxygen, but instead of having that attack the electrophile, it's going to come down and re-form the carbonyl, and that is going to push these electrons out.1871

We are going to form this bond here between the α carbon and the methyl group.1889

These two electrons are going to come and attack the carbon, and that is what kicks off the leaving group; so it is the α carbon that is nucleophilic and does the SN2.1894

So now, when we follow the arrows around, I see that I am back--my oxygen has two lone pairs; I have my carbonyl back; and attached to that α carbon, I have my alkyl group.1905

OK, this is typically how we are going to be drawing our enolates for the rest of the chapter, and how we can use them.1917

This is not wrong; this is still an acceptable drawing of an enolate; it is just not as good a representation as this one.1924

What else can we do with the α carbon?--we can also halogenate at the α carbon; that reaction is described as the α halogenation.1940

Here is an example: if I take this ketone and treat it with chlorine and base with water, I can install a chlorine in the α position.1947

Now, that mechanism is analogous to the one we just saw for α alkylation; we deprotonate the α carbon, and then we attack the Cl2 to put in the chlorine.1959

OK, and we can look at that reaction; but this reaction is not that useful, because this product that we formed--if you take a look at that α carbon, it turns out that this is now more acidic than your starting α carbon.1969

And why is that?--well, we just added a chlorine here; we know chlorine pulls electron density; or any halogen is electronegative and pulls electron density here; so that enhances the acidity here and makes it easier to deprotonate here.1987

So now, I am going to be forming some of this enolate, and I'm going to end up adding a second chlorine; so we could have multiple halogenations.2000

This would not be a very good way to make this 2-chloro, (1, 2, 3, 4, 5) 3-pentanone, because it would be hard to stop in this case.2010

Any time your product is more reactive than your starting material, it is difficult to control those reactions.2021

OK, in fact, it is so reliable that there is a test called the iodoform test that counts on this overreaction.2027

It is when you have a methyl ketone (so I have a CH3 here, and attached to a carbonyl, that is called a methyl ketone)--if you treat a methyl ketone with iodine and sodium hydroxide, what happens is: you end up replacing all three of those protons with iodines by successive α halogenations.2035

And that turns this group into somewhat of a leaving group; now, it's very unusual to have a carbon leaving group, but because that carbon has three iodines on it, all pulling electron density away, it is a reasonable leaving group.2058

And so, hydroxide can add into the carbonyl now and do a substitution reaction.2073

We can get a CTI, and again, even though we don't normally think of this carbon as a leaving group, it can, in fact, get kicked out; and the product we are going to get--we will get CI3-, which gets protonated to iodoform.2081

We will get the carboxylic acid product out here when we are done; and this ends up being a precipitate; this is a yellow precipitate, and this is the test.2098

The iodoform test is when you treat a sample with iodine and sodium hydroxide; if a precipitate forms, then that is a chemical test, a qualitative test, to indicate the presence of a methyl group attached to a carbonyl.2110

So, these kinds of tests are outdated, now that we have such powerful spectroscopic techniques, like NMR, where you can really see a methyl ketone a lot less ambiguously.2124

But this still has some historical note, and you will often see this reaction still, in textbooks.2135

With a base-promoted reaction, we expect to have multiple halogenations.2141

If we want to do a single halogenation, that is possible; but what we are going to do is: we are going to use acid-promoted reaction conditions instead.2147

In this case, if we used Br2, but with acetic acid now instead of a base (because here we see a carboxylic acid, so we recognize that these are acidic conditions), it is possible to get a single halogen in here with good yield.2157

OK, but because our conditions now are acid-catalyzed, rather than base-catalyzed, we can't use the enolate; we are going to use the enol instead.2174

The first two steps are going to be making the enol (remember, we already looked at that mechanism).2183

It might help to think about what the enol structure looks like; the enol has (instead of a carbonyl) an OH, and then it also has the double bond between what used to be the carbonyl carbon and the α carbon; so that is the enol.2193

This is our tautomerization mechanism; it's just a series of protonations and deprotonations; I'm going to take this α carbon; I'll use the A- that I just had, since it's catalytic in acid, because I use the acid in the first step, and I get it back in the second step.2230

I form the π bond and move the π bond; so 2 steps--we should always be able to form an enol.2245

Now, this product has both an aldehyde component and an alcohol component; so this is described as an aldol product, because it has an aldehyde and has an alcohol.2429

This is actually one of the few cases in organic chemistry where this is not named after Professor Aldol; this is not a named reaction that is capitalized; it is simply described as an aldol because that is what the product might look like.2442

OK, and we will also see that, if you heat an aldol product, you can cause it to lose water; so this aldol product will react further upon addition of heat.2456

It readily loses water to form this carbon-carbon double bond; so we'll look at both of those mechanisms, "How do we get to this aldol product?" and then "How does that aldol product lose water?"2474

OK, let's look at the base-catalyzed mechanism first: and when we are in base, we are going to deprotonate first--let's ask where would be a good place to deprotonate.2487

Certainly, it's going to be the α carbon, because that is the theme for this unit; and so, this is not an α carbon; this is not an α proton; this is attached directly to the carbonyl.2499

That is not at all acidic; but here is my α carbon; here is an α proton; let's change that to a CH2, so we can see one of those hydrogens.2511

And, absolutely, that is going to be our first step: to deprotonate that α carbon to form an enolate.2518

Now, every one of these steps is reversible, so we will use our equilibrium arrows here to go from one to the other.2525

OK, so my first step is to make an enolate, and what kind of reactivity do we expect for an enolate?2535

Well, it is a nucleophile; so we need to look around for an electrophile.2541

OK, and this is where it becomes important the choice of base that we use.2548

We only used hydroxide; we didn't use LDA in this case--we just used hydroxide; hydroxide is a weaker base, so it has taken some molecules of this aldehyde starting material and made the enolate.2552

But, by and large, most of this aldehyde is still the aldehyde; so the aldehyde is still present.2567

Therefore, it can serve as our electrophile; the carbonyl can be our electrophile.2575

So, we let our enolate do its thing: form the π bond; form the carbonyl; kick the π bond out; and attack the carbonyl; and then, what happens when a nucleophile attacks the carbonyl?2582

And now, typically, after we have made an enolate, that enolate that we have used thus far on mechanisms--that enolate has always formed the carbonyl and kicked this π bond out to some external electrophile.2779

But, in this case, what we have is a leaving group that is in the β position.2797

Remember, the α position is the first carbon next to the carbonyl; the β position is the next one over; and hydroxide can act as a leaving group.2805

So, when you have a leaving group in the β position, that enolate will re-form the carbonyl, shift this π bond down, and kick the leaving group off.2812

We are going to deprotonate as our first step; and step 2--we are going to inject the β leaving group.2823

And, when we do those three arrows, the product we get is the aldol product that we are expecting: the carbon-carbon double bond in between the α and the β carbon.2831

This is our mechanism: this step here kind of looks like a collapse of a CTI, a little bit.2846

Remember, collapse of a CTI was when we had an O- kicked down, and on the same carbon, had an OH; it would kick off the OH.2858

This is an extended system, but it is still the O- kicking down and forcing the OH off.2864

OK, now think about it: we just lost hydroxide here; that is not a typical leaving group, but we have seen it before for collapse of a CTI; so that is acceptable.2872

And now, we are seeing that, because this mechanism is similar, it is also acceptable for a β elimination mechanism.2883

You can lose hydroxide as a leaving group; OK, and remember, the reason for both of these--the reason that it is OK--is that your driving force is your formation of the carbonyl.2888

In both of those mechanisms, what is pushing the hydroxide out is an O-, which is forming a carbonyl at that same time, so because I am forming this carbonyl, it is OK to lose hydroxide as a leaving group.2907

OK, just a quick note of caution: this loss of water--this mechanism is not an E2 mechanism.2919

OK, it is very tempting here (I have redrawn it)--if I wanted to go from this aldol product and eliminate water, what is very tempting is that I just use my base to deprotonate.2928

And, instead of using it to form an enolate, like I need to, I think, "Well, let's just form this π bond and kick out the leaving group; wouldn't that be a fast, easy mechanism--shouldn't that be better?"2940

OK, that would be an E2, when we have a single-step elimination mechanism; that is described as E2.2951

OK, but that does not happen: there is no way we can do an E2, because hydroxide is not an acceptable leaving group for such a mechanism: we have to have a traditional leaving group like bromide, chloride, iodide, tosylate, something like that, in order to enable this.2958

The only reason this elimination can take place is because of this carbonyl, because you have this resonance-stabilized enolate intermediate, and so it is critical and must be used as part of your mechanism.2975

OK, so just a reminder not to do a single-step elimination here of water, but to do a two-step β elimination.2985

OK, so let's summarize the aldol: an aldol is what we have when we have an aldehyde or ketone; we are going to treat it with either base or acid; the ketone (in this case) is going to serve both as the nucleophile (the α carbon is doing the nucleophile), and it is going to serve as the electrophile (the carbonyl of the second molecule is going to be the electrophile, so there is the bond we are forming).2996

We form a new carbon-carbon bond between the α carbon in one and the carbonyl carbon in the other.3020

OK, so the product we get could be described as a β hydroxy ketone, right?--not in the α position, but in the β position, where it's going to have this hydroxy group.3027

And then, if we heat this, we can lose water, and so now we form a double bond between the α carbon and the β carbon; so this structure is described as an α β unsaturated ketone, because we have a point of unsaturation; we have a double bond between the α carbon and the β carbon.3039

So, when we do an aldol, we can kind of choose which kind of product we want to draw: it's either the β hydroxy ketone, or sometimes it's an α β unsaturated ketone; very often, the trigger on whether or not you are going to lose water will be the addition of heat.3061

OK, sometimes this is spontaneous, because this dehydration is a relatively easy one.3076

OK, let's think about why that is: typically, when we dehydrate an alcohol, we need strong conditions: sulfuric acid, phosphoric acid, heat--it's very hard to dehydrate an alcohol.3085

This one happens very, very easily: you don't need a strong acid; you could even do it in base; and you just need to warm it up a little.3095

OK, but think about your final product: you are not just forming an ordinary carbonyl; it is not just an ordinary alcohol; it is a β-hydroxy alcohol...it is β to a carbonyl, and so this resulting double bond is now going to be conjugated with the initial carbonyls.3106

This is a conjugated system, and that conjugation--when you have a double bond right next to another double bond, that means you have resonance, and it is a very stable structure.3128

So, that is why aldol products are very likely to lose water and go to an α-β unsaturated carbonyl.3144

OK, so so far, we took a look at the base-catalyzed mechanism; we will look at the acid next, but let's draw some comparisons.3155

In the base-catalyzed mechanism, everything was either...all the species were either neutral, or they had a negative charge; the nucleophile we used was the enolate; the electrophile we used was the ketone carbonyl.3162

OK, now how is that going to change for an acid-catalyzed mechanism?3182

Now, an acid-catalyzed mechanism--you can't have a strongly basic species, like an enolate.3185

You need to have either neutral compounds or positive charges: you are not going to have any O- in an acid.3193

So, instead of the enolate, what species could we use that would still be nucleophilic?3198

How about if we protonated that to make it an OH; what would that be called?3204

That is the enol, and that is going to be the nucleophilic species in the acid-catalyzed.3209

We also can't have a neutral carbonyl, because if you attack a neutral carbonyl, that forms an O-; you can't have an O- in an acid-catalyzed mechanism, so what we are going to do is: we are also going to protonate the carbonyl, and it will be a protonated carbonyl that gets attacked.3214

So, it turns out that this mechanism is going to be a little longer, in order to accommodate these species; but let's take a look at that.3232

We said we had to make an enol; those will be our first two steps of the mechanism, then: we will start with our...we are doing acetone, in this case; and we have some acid present.3244

So, to make an enol, first we protonate; and then the second step to make the enol is deprotonate; so we take this α proton, A-, and we have an enol.3256

Now, I have my nucleophile: what electrophile is it going to react with?3288

Well, we just saw on the previous slide: it is not going to react with a neutral carbonyl; it needs to react with a protonated carbonyl.3292

We just showed the mechanism on how to form the protonated carbonyl, so we can use that again; but this is going to be our electrophile.3300

What is the bond that we are going to be forming?--it's going to be between this α carbon and this carbonyl carbon, so our mechanism to get there is: our enol starts at the lone pair on the oxygen, forms the carbonyl, and that is what kicks the π bond out.3308

It kicks those electrons out; those attack the carbonyl of the electrophile and break the π bond; that is the key step in any aldol--when the α carbon attacks the carbonyl carbon.3327

What do we have to do to finish this up?--it looks like we just have to deprotonate; so we'll bring A- back in here, and we are done.3368

There is our aldehyde; so once we made our enol, then we protonated the carbonyl; we attacked the carbonyl; and we deprotonated the carbonyl.3385

OK, but it is going to be a little bit longer mechanism to get where we want to go.3395

Notice, overall there are no O- charges in the acid; everything is positively charged or neutral--that is going to be consistent with any acid-catalyzed mechanism.3397

Now, let's think about losing water from that aldol product to form the alpha, beta unsaturated ketone; we describe that mechanism as a beta elimination mechanism; we saw, in base, it was just two steps.3416

So, how are we going to adjust this for the acid-catalyzed version?--we can't have an enolate, so instead, we are going to make the enol.3446

OK, and what do you think about hydroxide--do you think that would be an acceptable leaving group in acidic conditions?3455

No way--much too strong of a base; so what would be a leaving group that would be OK to have around in an acidic solution?3460

We would have to lose water; OK, so with that in mind, let's think about our mechanism: let's give it a try.3468

We are going to start with our aldol product; we want to lose water from this, so the first thing we need to do is: we need to convert this to the enol.3475

How do we make the enol?--two-step tautomerization mechanism--2 steps.3489

First, we protonate; that tautomerization practice is really going to pay off when you get to these more complex mechanisms, so that hopefully this part will be somewhat automatic: protonate at the carbonyl, and then deprotonate at the α carbon.3495

That is the first thing we have to accomplish: we have to make the enol.3520

OK, and now, we need to get rid of our beta leaving group, but it is not a good leaving group right now.3527

If this kicked down and kicked off, the leaving group would be hydroxide leaving.3532

So, how do we make it a better leaving group?--we need to protonate, and so now it is going to look like water when it leaves; it looks like water right now, as it's attached.3535

This is a good leaving group in acid; so when we said that hydroxide is an OK leaving group for collapse of a CTI for beta elimination, that is true, but that assumes you are looking at a base-catalyzed mechanism, because you have a hydroxide.3554

In this case, we need water as our leaving group; so now, we are ready to do our beta elimination and eject that beta leaving group.3569

The same idea as with the enolate, though: we are going to start up here in the lone pair on this oxygen and form the carbonyl; that is what pushes this π bond down; that is what kicks our leaving group off.3577

A-...of course, that is the only negative charge we can have in an acid-catalyzed mechanism: A- represents some very stable conjugate base of a strong acid (like sulfuric acid conjugate base, let's say).3611

So again, it was just two steps in base and several steps in acid; but practice with the base--get used to that; understand the logic, and be able to talk yourself through the mechanism, and know where to go, so that you can make those adjustments and make it work for acid.3634

Now, in all the cases we have looked at so far, aldol has always been a self aldol, meaning one molecule of a ketone has reacted with the exact same structure of another molecule of a ketone.3657

OK, when we have two different ketones or aldehydes coming together, this is something that can be reasonable, if only one of the compounds has α protons.3670

So, for example, if we bring it together--acetone and benzaldehyde with base (sodium hydroxide and water), we think about who could be the nucleophile and who could be the electrophile.3681

Well, we know we have a base here; so we can deprotonate, and we are going to look to our α carbons to deprotonate.3693

Well, acetone certainly has an α carbon; so this one--when we deprotonate it, that would be our nucleophile.3703

We have one nucleophile here, but what about benzaldehyde?--there are no α protons on the benzene ring, and this hydrogen is not alpha, so this has no α hydrogens; so that means this cannot be a nucleophile in any way.3708

We only have one nucleophile; how about looking for electrophiles?3728

Well, both of these structures have carbonyls, so you might think they both could be the electrophile in the reaction.3732

OK, however, we have a ketone and an aldehyde, and the aldehyde is the better electrophile.3738

Now, let's just take a quick look at why that is: OK, well, remember: alkyl groups donate electron density; so when we look at a ketone, we know that these carbons are going to be adding electron density to the carbonyl.3750

OK, is that good for an electrophile?--that is not good for an electrophile.3765

OK, this is more electron rich; so this is a poorer electrophile, and the aldehyde is the better electrophile, because this only has one alkyl group donating; this has a larger partial positive.3770

So, comparing an aldehyde and a ketone, an aldehyde is going to win, in terms of electrophilicity and reactivity.3784

And so, comparing acetone and benzaldehyde, benzaldehyde is a better electrophile.3791

We have one electrophile; we have one nucleophile; which means we can expect to get one major aldol product--and what would that look like?3798

Well, we could start by drawing our acetone: we know we are going to form a new bond between...this α carbon is our nucleophile, and where is it going to attack?3809

It is going to attack this benzaldehyde carbonyl; so this used to be a CH3; it is now a CH2, because we deprotonated there, and that is how we have room for this bond.3818

And then, when we are all done, what do we have on this carbon?--we have a phenyl, we have a hydrogen, and we have an OH.3830

It would be an O-, but we had protonated at some point, so we get this β-hydroxy ketone; this would be our major product.3837

However, this probably would lose water spontaneously; we wouldn't really have to warm this up, probably; just having it at room temperature would be enough for this to lose water.3845

In this case, it is very likely that you would get this as your major product.3860

OK, and let's think about this particular example: why is it, in this case, that it would be very difficult to isolate this alcohol--very difficult to not have it undergo this dehydration step, this β elimination step?3868

Well, that is because we have...this double bond is not only in conjugation with the carbonyl, like it always is in an aldol; but we also have this benzene ring right here, so this is conjugated with the benzene ring π bonds, and that makes this π system an extended, conjugated π system--very, very stable.3881

So, in the case with benzaldehyde, it's very common to seemingly spontaneously lose water and go straight to the alpha, beta unsaturated carbonyl, even though you haven't seen the addition of heat explicitly.3905

Now, what if I wanted to do a mixed aldol between two ketones or two aldehydes--two groups that have similar reactivities?3923

Well, we could exert control there; we wouldn't just want to mix those two together in the presence of base, because then, you can get all sorts of possible products out.3931

You have two possible nucleophiles; you have two possible electrophiles; so in a mixed aldol like that, you might have up to four possible products.3939

But the way you could exert control is: instead of just mixing everything together, you could do it step-wise by using something like LDA.3947

Remember, LDA is our strong base; and so, if we started out with just a single ketone, and treated it with LDA, we would expect that ketone to be deprotonated.3954

Now, this is an interesting case, because we have two different α protons in this case; how would we know which one is the major site of deprotonation?3967

Well, remember: lithium diisopropylamide had those two isopropyl groups; it is a very bulky base; LDA prefers to attack the less hindered α carbon (and therefore, α proton).3978

So, it has to do with sterics; it has to do with kinetics; it goes to the less hindered side.3995

So this is going to deprotonate to give the following enolate: this is a CH2...so we have just one nucleophile; we have this enolate being formed.4001

Remember, the reason we use LDA--we use a full equivalent of LDA so this gets completely converted to this enolate.4021

OK, and now, after the enolate is formed, now we add in a second carbonyl; we have just one electrophile, so there is really just one major product that can occur.4028

What is the bond that we are going to form?--the electrophile has reacted with the carbonyl carbon; the enolate is reactive at the α carbon; and so, we expect the enolate to attack the carbonyl.4039

We are going to form a new bond to the α carbon, like we would in any aldol; this carbon is part of the 6-membered ring; and then, on this carbon, we also have a single bond O-, which then gets protonated upon workup.4059

Now, we would do a separate workup: when our reaction is all done, we get this β-hydroxy carbonyl.4076

Now, I want to point out just some stuff to you, some common mistakes: it is very common to draw as the product of this, this product; and what would be the problem with drawing this product?4084

You have attached the carbonyl carbon to the carbonyl carbon; something is missing here.4098

OK, you have an α-hydroxy carbonyl; that is not the product you should be getting--it should be a β-hydroxy carbonyl.4105

So, especially when it comes to line drawings, it is very easy to lose carbons when you are drawing your zigs and zags; so make sure you explicitly show your α carbon that was your nucleophile.4111

There is a new bond to your β carbon, which was your electrophile (that carbonyl carbon).4121

And that way, you won't make mistakes of accidentally losing carbons when you are dealing with line drawings.4126

Let's see if we can take one of these problems and do it as a retrosynthesis: if we are given this starting material...this target molecule, rather...and asked to synthesize it, how could we do that?4136

Can we predict what reagents we need for the aldol to give this target molecule?4149

Well, it looks like an aldol product, even if it doesn't give us that clue; it looks like an aldol product, because between the α and the β carbon, we have a double bond.4167

This is an alpha, beta unsaturated carbonyl (ketone, in this case).4176

And that is the type of functional group pattern we would get if we did an aldol reaction.4185

So, it is possible to just kind of disconnect the bonds that we know that are being formed; but I think maybe an easier step backward would be to say, "OK, well, remember: I got this double bond because I lost water from the β-hydroxy ketone."4192

Maybe you can kind of add water back in; let's abbreviate this as just a phenyl group; let's add water back in, so that we add the hydrogen, and we add OH.4208

We already had hydrogens here, so we can leave those in if we want, so we don't get too confused.4221

Let's add water back in to go back to the β-hydroxy ketone, because this is also a pattern that we could get from the aldol reaction, but it's a little easier to grab onto this one, because we have some more interesting functional groups.4225

OK, the key disconnection we make: remember, what is an aldol?--it is the reaction of an α carbon to a carbonyl carbon.4243

This is the key disconnection that we make in an aldol; it is between the α and the β carbons.4251

This carbon--remember, one of these carbons was a nucleophile; one of them was an electrophile; that is how they were able to come together to make this product.4256

The α carbon, of course, was the nucleophile; this α carbon was the nucleophile--in other words, that was the enolate in the mechanism.4267

And that means that this carbon was my electrophile; what electrophile could I have that, after it gets attacked, we end up with an alcohol at that carbon?4280

It was a carbonyl; if I had a carbonyl here, and an enolate attacked it, it would break the π bond, and I would get an alcohol there.4292

That is the logic we can use: when we do this disconnection, then, we end up with just a phenyl and a 1-carbon aldehyde (it looks like benzaldehyde, again, is involved in this).4302

That would be our electrophile, and what would our nucleophile be?--it would be a phenyl group with a carbonyl and then a CH3.4313

So, if I had acetophenone and benzaldehyde, I could actually just mix these two, because this is similar to the mixed aldol we just looked at--because this is a ketone and this is an aldehyde.4328

I could just mix these two together in the presence of base (sodium hydroxide and water, or methoxide and methanol--something like that), and a little heat to afford the dehydration.4341

These actually would come together to form this single product as our target molecule, what we expect to get.4352

This is our major aldol; we wouldn't have to bother first treating this with LDA and then adding the benzaldehyde separately.4359

OK, but this would be a good process by which we could come up with the starting materials required for an aldol synthesis.4366

Next, let's take a look at a reaction called the Claisen condensation: this is very much related to the aldol condensation; in fact, it is an aldol reaction; but instead of using a ketone or an aldehyde, we are going to be using an ester instead.4376

We have a base, so let's deprotonate something; where do we deprotonate?--α carbon.4415

α carbon, α carbon, α carbon: that is the whole theme of this lesson.4420

So, I am going to look: and remember, these carbons are not attached to the carbonyl; those are not α carbons, but this one is; so I could use ethoxide as my base, and I could deprotonate the α carbon.4425

This is done reversibly, because ethoxide is not LDA--it doesn't do so in just one direction.4439

But I will get out some of this enolate; let's just abbreviate this--this is an ethyl group here, so we can abbreviate, OEt, and save us a little time drawing it.4448

OK, so deprotonate to form an enolate: that is a good idea; and what kind of reactivity does an enolate have?4461

We know that is a nucleophile; so let's look around for potential electrophiles.4468

Well, remember: because we use this weaker base, ethoxide, we still have most of this ethyl acetate, still as the ester form; so that is still around.4474

We can bring in another molecule of the ethyl acetate as our electrophile; so see how it is just like an aldol?4486

We have an enolate of a carbonyl-containing compound attacking a non-deprotonated copy of itself; so our enolate does its thing; re-form the carbonyl; break the π bond.4493

And then, what happens when you attack a carbonyl?--always, always, always the same: break the π bond and move it up on oxygen.4507

Let's draw this product over here; let's draw our enolate again--it now has a carbonyl.4516

Single bond to the α carbon; new carbon-carbon bond here; very good, and then we have...what is on this carbon?...we have a CH3; we have an O-; and we have an O ethoxy--we have an O ethyl.4526

OK, now if this were an aldol reaction, if we were using aldehydes and ketones, our reaction would be done right now...essentially--almost done.4547

We deprotonate; we attacked; and then, all we would do is reprotonate; we would have our β-hydroxy ketone product.4554

OK, however, when you are dealing with an ester, it takes a different path, because what happens when a nucleophile attacks an ester?4562

You now have an O- on the same carbon as a leaving group; we call this a CTI (a charged tetrahedral intermediate).4571

So, this is not a stable intermediate: this is not just sit-around-and wait-for-something-to-come-and-protonate-it.4581

What happens to a CTI?--we use two arrows, and we collapse that CTI.4590

All right, what happens with esters when they get attacked by nucleophiles?--we get addition-elimination; we get a substitution at that carbonyl.4615

So, taking the place of this ethoxy group is going to be the enolate group.4621

OK, and it looks like our reaction is done here, because we are at a nice, stable, neutral, happy product; and this is the ultimate product that we will be isolating.4630

However, in the reaction conditions, this product is not stable, because our reaction conditions are basic.4640

And what do you see about this product that makes it want to react with base?4648

We have an active methylene here; we have an active methylene that has two EWGs, two electron withdrawing groups; it has an ester and a ketone attached to it; that makes it very acidic, and because we are in basic conditions, we are going to deprotonate this.4653

OK, to make this stabilized enolate, of course, this has lots of resonance; so this is very, very stable; and in fact, this last step--this final deprotonation--is the driving force for the Claisen.4692

If we did not have this last step--if, for some reason, you didn't have a hydrogen here, and you couldn't deprotonate--the Claisen actually wouldn't go; the retro-Claisen would happen instead.4708

It is because this final step is essentially irreversible--this is such an acidic proton--that we get this as our final product.4722

OK, so that is what happens after step 1; and now you can see, perhaps, why we need this step 2: we need to have an acidic workup to reprotonate this enolate that has formed as the final product, and that is how we can get the product that we kind of imagined we were going to get.4732

We can go back to this neutral intermediate that we had for a fleeting instant, and we come back to our actual Claisen product.4753

So, how do we describe the Claisen product?--it is described as a β-keto ester.4762

We have added a new group; we have added a new bond to our α carbon; and the group that we have added is another carbonyl, because we had addition-elimination at that ester group.4770

We had a substitution there; so that regenerates the carbonyl in the β position, so we get a β-keto ester when we do a Claisen condensation.4781

Let's take a look at an example and see if we can predict the product here.4794

OK, we have, in this case...we have a methyl ester; so that is why we are using sodium methoxide and methanol.4798

Remember, any time we want to have a base around an ester, we want to use that matching base.4804

And so, what do you think would happen here?--well, I think this base is going to deprotonate our α carbon.4812

So, I think we are going to have this anion as our nucleophile (I'm just seeing if we can do a little work here, without going through a complete mechanism; let's see if we could come up with the product).4822

What is that going to react with?--well, there are no other ingredients in here, so this is going to just be a self Claisen condensation; we are going to have another molecule of our ester.4834

We can draw the carbonyl down here, if you would like, so it's a little easier for it to attack.4848

I'll bring a second structure in here, and that is our electrophile; our key bond, then--just like the aldol: the α carbon of one to the carbonyl carbon of the next.4856

But this is going to attack the carbon and break the carbonyl π bond; but then, what is going to happen next?4868

That O-, in the following step, is going to re-form the carbonyl and kick out the ester.4875

OK, so it is not an SN2 mechanism; you don't want to just bring the arrow in and show it kicking out--that implies a back side attack; but even if you want to do some of this work in your head, you can imagine it going up into the carbonyl and then coming back down out of the carbonyl and kicking off the leaving group.4881

OK, well, any time you are reacting something with base, that means we are going to lose a proton; we are going to deprotonate somewhere, and we are going to lose what kind of proton?--we are going to lose an α proton.4984

So, this is a symmetrical molecule: it doesn't matter which side we deprotonate; let's deprotonate over here.5001

We could do that; OK, there is our α carbon--that is our nucleophile; and now, rather than having a second molecule that we can attack, because this has two functional groups, let's see if possibly we can get an intramolecular reaction to take place.5010

Do we have any electrophile, internally, that is the appropriate distance away?5025

So, let's check: if this is our first carbon, this is 1, 2, 3, 4, 5 atoms away from a carbonyl which is electrophilic.5031

Would forming a 5-membered ring be a favorable ring to form?--absolutely: 5- and 6-membered rings are the ones that we are always going to be looking for.5044

So, what we could do is: we could draw a nice 5-membered ring: 1, 2, 3, 4, 5, and number it: 1, 2, 3, 4, 5, and then just add in our groups on what is missing.5054

On carbon 1, what do we have?--attached to that is where we have our ethyl ester, because carbon 1 was the α carbon, so it should be α to a carbonyl.5066

And then, we had a CH2, a CH2, CH2, and then carbon 5 has what?--it has the new bond to carbon 1, of course.5078

Here we can show it as the bond that we are imagining forming; so it has the new bond to carbon 1, and what else does carbon 5 have?5086

That is what we would get as one of our intermediates, after we deprotonated and attacked.5107

All right, we can imagine this attacking and breaking the π bond up; and where do we go from here?--well, this looks like a charged tetrahedral intermediate, and so we are going to have this collapse with our two arrows.5113

Our O- comes down and kicks off our leaving group, and we end up forming a ring that still has that same pattern of a beta-keto ester that we expect for a Claisen.5127

An intramolecular Claisen condensation has its own name: it is called a Dieckmann reaction; but is the same exact sort of reaction.5142

Aldol reactions can happen intramolecularly, as well, to form rings, and a Dieckmann is when we have a diester cyclizing to do an aldol-type condensation.5151

This wraps it up for part 1 of our enols and enolates discussion, and we will continue the discussion, looking at some different types of species, different types of electrophiles, that enols and enolates can react with, next.5163

Related Books

This is the actual book that Dr. Laurie Starkey teaches out of as a professor, and manages to be both clear and yet precise. The book presents a logical, systematic approach to understanding the principles of organic reactivity and the mechanisms of organic reactions.

This book includes all of the concepts typically covered in an organic chemistry textbook, but special emphasis is placed on skills development to support these concepts. This emphasis upon skills development will provide students with a greater opportunity to develop proficiency in the key skills necessary to succeed in organic chemistry.